Quantitative Aptitude

Permutation and Combination

609 Questions
All Vowels or Objects Together 139 Restricted Group Selection 135 Objects Never Together 98 Fundamental Counting Principle 96 Objects with Repetitions Arrangement 72 Rank or Dictionary Word 20 Selection With Identical Items 15 Pairing or Grouping 10 Distributing Distinct Objects 9 Distributing Identical Objects 9 Circular Ring Arrangement 5 Circular Necklace Arrangement 1

Permutation and Combination Questions

Multiple choice

A test has $4$ parts, the first $3$ parts carry $10$ marks each and the $4^{th}$ part carries $15$ marks. Assuming that marks are not given in the form $p/q\space\left ( p< q, p,q\in I \right ), q\neq 0$ then the number of ways in which the student can get $30$ marks out of $45$ is
  1. $5300$
  2. $711$
  3. $6011$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a partition problem. We need to find non-negative integer solutions to x1 + x2 + x3 + x4 = 30, with constraints 0 <= xi <= 10 for i=1,2,3 and 0 <= x4 <= 15. The coefficient of x^30 in (1+x+...+x^10)^3 * (1+x+...+x^15) is 711.

Multiple choice

Directions: The following question contains 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of worlds that appear before the word COCHIN is

  1. 360

  2. 192

  3. 96

  4. 48
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The letters of COCHIN contain C twice. Before the second letter O, the remaining letters C, H, I, and N can occur, giving 4 x 4! = 96 permutations.

Multiple choice

A TCYonline test contains 15 different questions with one question being the easiest and one being the toughest. In how many ways can these questions be arranged, one after the other, so that the easiest and the toughest question never appear together?

  1. 14! – 2!

  2. 15! – 2!

  3. 15! – (2!13!)

  4. 15! – (2!14!)

  5. 14!
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total arrangements of 15 questions is 15!. Arrangements where the two specific questions are together: treat them as one unit, so 14! * 2!. Thus, arrangements where they are NOT together = 15! - (2! * 14!).

Multiple choice

An examination contains 3 sections, each having 6 questions. In how many ways a student can attempt 12 questions such that in each section he has to attempt at least 3 questions and at the most 5 questions?

  1. 5,400

  2. 3,375

  3. 8,775

  4. 14,175

  5. 6,175
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We need to choose 12 questions from 18 (3 sections of 6). Constraints: 3 <= n_i <= 5 for each section i, and sum(n_i) = 12. Possible distributions (n1, n2, n3): (4,4,4), (3,4,5) in 6 permutations, (3,3,6) is invalid. Calculating combinations: C(6,4)^3 + 6 * C(6,3)*C(6,4)*C(6,5) = 15^3 + 6 * (20 * 15 * 6) = 3375 + 10800 = 14175.

Multiple choice

From a group of six persons A, B, C, D, E and F, a committee of four is to be formed. In how many ways can it be done if E is also selected whenever B is selected?

  1. 5

  2. 6

  3. 0

  4. 11
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total ways to choose 4 from 6 is 6C4 = 15. Cases where B is selected: B must be in, E must be in. We need 2 more from the remaining 4 (A, C, D, F), which is 4C2 = 6. Cases where B is NOT selected: We choose 4 from the remaining 5 (A, C, D, E, F), which is 5C4 = 5. Total valid ways = 6 + 5 = 11.

Multiple choice

In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?

  1. 6!/2!

  2. 3!*3!

  3. (3!*3!)/2!

  4. (4!*3!)/2!

  5. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The word ABACUS has 6 letters: A, B, A, C, U, S. Vowels are A, A, U. Treat (AAU) as one unit. We have 4 units: (AAU), B, C, S. These can be arranged in 4! ways. Within the unit (AAU), the letters can be arranged in 3!/2! ways. Total = 4! * (3!/2!) = 24 * 3 = 72. Option D matches this expression.

Multiple choice

Fifteen students are to be placed in a queue. Out of these, 3 always stand together and out of the remaining 12, 2 do not want to stand together. Find the number of possible arrangements.

  1. P(12, 3) x 2!

  2. 12!3! x 11

  3. 12!3!2!

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Treat the 3 students as one block. We have 13 items (1 block + 12 others). The 2 students who don't want to be together can be handled by total arrangements minus arrangements where they are together. The calculation results in 12! * 3! * 11.